[PDF] - CSC 473 Automata, Grammars & Languages 9/29/10 Automata, PDF Document

SLIDE 1

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 1

C SC 473 Automata, Grammars & Languages

Automata, Grammars and Languages

Discourse 03 Finite Automata

C SC 473 Automata, Grammars & Languages 2

Finite Automata / Switching Theory

(CS) / (CE)

Boolean operators / Gates (Elem. Switching Ops)

1 1 1 1 1 1 1 1 1 1 1 ¬ x x ⊕ y x ∨ y x ∧ y y x

C SC 473 Automata, Grammars & Languages 3

Boolean Functions / Combinatorial Circuits

Circuit

∨ ∧ ∧ ¬

1

x

2

x

1 (sum)

z

2 (carry)

z

1 2

x x

1

2

x x

1

2

x x

1

2

x x

H half adder

H H

1

x

2

x

∨

3

x

1 2

x x

1 (sum)

z

1 2

x x

1

2 3

( ) x x x

2 (new carry)

z

F full adder

(old carry)

SLIDE 2

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 2

C SC 473 Automata, Grammars & Languages 4

Boolean Functions / Comb. Circuits (contʼd)

Table representing F

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

z2 z1 x3 x2 x1

C SC 473 Automata, Grammars & Languages 5

Boolean Functions / Comb. Circuits (contʼd)

Equations representing F
General scheme (n inputs, m outputs)

1 1 1 2 3 2 2 1 2 3 1 2

( ) ( ) (( ) ) ( ) z g x x x x z g x x x x x x = =

=

=

1

2 1 2 1 1 2 1 2

( ) ( , , , ) ( , , , ) ( ( , , , ), , ( , , , ))

m n n m n

z g x z z z g x x x g x x x g x x x = = = … … … … …

C SC 473 Automata, Grammars & Languages 6

Finite Automata / Sequential Circuits

Add “memory” elements = delay elements
Finite # of delay elements possible ⇒ ∃ d

( ) y i ( ) ( ( ), ( 1 ), ( 2 ), ) z i f x i x i x i =

…

( 1 ) y i + x z

f

combinatorial circuit

( ) ( ( ), ( 1 ), ( )) z i f x i x i x i d =

…

SLIDE 3

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 3

C SC 473 Automata, Grammars & Languages 7

Finite Automata / Sequential Circuits

Ex: sequential adder: add 2 binary numbers; low
rder bits received first
(a) sequential net (circuit):

1001 0101 1110 carry

1

( ) x i

2

( ) x i

1

( ) z i

full adder F

1

( 1 ) y i +

1

( ) y i

1 2

( ) ( ) x x

C SC 473 Automata, Grammars & Languages 8

Finite Automata / Sequential Circuits

(b) Next-State & Output Equations:
(c) Transition Table: state space

1 1 2 1 1 2 1 1 2 1

( 1 ) (( ( ) ( )) ( )) ( ( ) ( )) ( ) ( ) ( ) ( ) y i x i x i y i x i x i z i x i x i y i + =

=
1

{ ,1 } { , } q q = = input alphabet ={(00),(01),(10),(11)}

utput alphabet ={0,1}
(

00 ) ( 01 ) ( 10 ) ( 11 ) q

1

q

0 / 0

q

next state/output table

q

0 / 1

q

0 / 1

q

0 / 1

q

1 / 0

q

1 / 0

q

1 / 0

q

1 / 1

q

x

C SC 473 Automata, Grammars & Languages 9

Finite Automata / Sequential Circuits

(d) State Diagram:

( 00 )/ 1 ( 00 )/ 0 ( 01 )/ 1 ( 10 )/ 1

q

1

q

( 11 )/ 0 ( 11 )/ 1 ( 01 )/ 0 ( 10 )/ 0

SLIDE 4

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 4

C SC 473 Automata, Grammars & Languages 10

Finite Automata / Sequential Circuits

(e) Finite-State Transducer (Mealy Machine)

 A 5-tuple

where

( , , , , ) M Q q

=

1 1 1 1

{ , } {( 00 ),( 01 ),( 10 ),( 11 )} { ,1 } : ( ,( 00 )) ( ,0 ) ( ,( 01 )) ( ,1 ) ( ,( 10 )) ( ,1 ) ( ,( 11 )) ( ,0 ) ( ,( 01 )) ( ,0 ) . Q q q q Q Q q q q q q q q q q q etc

=

= = = = = = =

finite set of states start state input alphabet

utput alphabet

transition/output function

C SC 473 Automata, Grammars & Languages 11

General Sequential Network

1 n

x x

1

m

z z

1

s

y y

…

:

n s m s

B B

+

+

{

,1 } B =

state space =

s

B

( ( ), ( )) ( ( 1 ), ( )) ( , ) ( , ) y i x i y i z i q a q b

=

+

=

δ is a boolean functio n

C SC 473 Automata, Grammars & Languages 12

Three Types of Automata

M M M

transducer recognizer (acceptor) Enumerator (generator)

3 2 1 n

a a a a

3

2 1 n

c c c c

time

3 2 1 n

a a a a

3

2 1 n

c c c c

i

a

Yes(1) No(0)

SLIDE 5

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 5

C SC 473 Automata, Grammars & Languages 13

Machines that Recognize

Detection of an “event”, i.e., a pattern in input
Recognition of just those words in some language L
Definition of a language
Ex: detect abab –all non-overlapping occurrences

a b a b b a a b a b

C SC 473 Automata, Grammars & Languages 14

Ex: C Comments /* … */

Filter in the lexical scanner
Recognizer

{,/} a

=

: aa

:

/ :

/

:

:

/ a a

:

/

: /

:

empty

/ :

:

a

:

:

a / a

a
/
/

(transducer)

notation in : out

C SC 473 Automata, Grammars & Languages 15

Finite Automaton (Finite State Machine, FSA)

Defn 1.5: A (deterministic) finite automaton is a 5-tuple
is a finite set, the states
is a finite set, the alphabet
is the transition function
is the start state
is the set of accepting (final) states
Ex:

( , , , )

,

M Q q

F

=
F

Q

Q

: Q Q

q

Q

( , , ,

)

,

M Q q

F

=
1

2

{ , , } Q q q q =

1

{ , } { } a b F q = =

1 2 1 2 1 2 2 2 2

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) q a q q b q q a q q b q q a q q b q

=

= = = = =

q

1

q

2

q

a b b a , a b

M

SLIDE 6

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 6

C SC 473 Automata, Grammars & Languages 16

How FA Compute

FA is a finite structure—like

a program—fixed and static

Need to define the behavior of M on input w

 Sequence of configurations  Like trace of a program on given data  Dynamic and input-dependent

Ex: start on input Look at sequence
f “moves” determined by the transition function:

Since in accepting state when input exhausted, w is recognized by

( , , , )

,

M Q q

F

=
M

w ababa = ( q0,ababa)( q1,baba) ( q0,aba)

(

q1,ba) ( q0,a)

(

q1,) ( q0,ababa)( q1,)

M

C SC 473 Automata, Grammars & Languages

www.jflap.org

JFLAP is a package of graphical tools which can be used as an aid in learning the basic concepts of Formal Languages and Automata Theory.

C SC 473 Automata, Grammars & Languages 18

Info on JFLAP

Website http://www.jflap.org/



Downloads



Tutorial

Lectura (linux) install



cd /usr/local/jflap



java -jar JFLAP.jar

X11 forwarding (graphics)



ssh -X lectura

SLIDE 7

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 7

C SC 473 Automata, Grammars & Languages 19

How FA Compute (contʼd)

Given a FA
Defn: configuration of M is an element of
Defn: yields in one step (or moves) relation

between configurations is defined by where

 Notes: is a function, since δ is. Is undefined.

Defn: yields is the relation

 Means “moves in zero or more steps to” 

Defn: A string w is recognized (accepted) by M ⇔

( , , , )

,

M Q q

F

=
(

q,aw) ( q ,w) ( q,a) =

q

Q

!

, , , a w q q Q

(

q,)

(

q,w) ( q,w) ( f F)( q0,w) ( f,)

C SC 473 Automata, Grammars & Languages 20

How FA Compute (contʼd)

Defn: The language recognized (accepted) by M is
Defn 1.16: A language S is regular iff there is some FA

that recognizes it, i.e.,

Ex: In FA

( )[ is a FA ( )] M M S L M

=

L (M) = { w : M accepts w} = { w :( f F)( q0,w)

(

f,)

}

( q0,a) ( q1,) ( q0,aba)

(

q1,)

(

q0, ( ab)

ia)(

q1,) L (M0) = {( ab)

ia : i 0

}

M

C SC 473 Automata, Grammars & Languages 21

Example:

Coin checker for 30¢ coffee. Σ={n,d,q}

15 5 10 30 25 20 35 40 45 50

i

= make change for i – 30 ¢ & vend coffee q n n n n n n d d d d d d q q q q q

SLIDE 8

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 8

C SC 473 Automata, Grammars & Languages 22

Regular Operations & Regular Expressions

The regular operations on languages are:

 union (∪), concatenation (⋅ ) and Kleene star (*).

So called because the class of regular languages are

closed under them—i.e., applying these operators to regular languages results in a regular language. (We will prove these closure results later.)

In fact, these three operations (∪, ⋅ ,*) actually characterize

what it means to be a regular language: any regular language can be built up from alphabet symbols and a finite number of these regular operations.

This motivates the notion of regular expression: a

sequence of symbols, like an arithmetic expression, that defines a regular language using regular ops.

C SC 473 Automata, Grammars & Languages 23

Regular Expressions

A syntax for describing sets of strings (languages)

 Terse  Eliminates fussy “{ }”  Reminiscent of arithmetic expressions  Obeys some useful “algebra”, e.g., (E*F*)* = (E+F)*

Syntax for regular expressions over Σ,+, ⋅ ,*,(,)

 E → (E+E) ( text uses ∪ not +; some authors use |)  → (EE)

(usually suppress the ⋅ in E ⋅ E)

 → (E*)  → ε (some authors use λ )  → ∅  → a for each a in Σ

suppress (,) where possible: (a+b)*a not ( ( (a + b)* ) ⋅ a )

C SC 473 Automata, Grammars & Languages 24

Regular Expressions (contʼd)

Meaning rules for the syntax
The meaning (denotation) of an expression, L(E), is a set
f strings (a language)
Rules

expression E language L(E) ∅ a {a} ε {ε} (E+F) L(E)∪L(F) (EF) L(E)⋅L(F) (E) L(E)

{} =

SLIDE 9

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 9

C SC 473 Automata, Grammars & Languages 25

Reg. Expr.: Examples, Equivalence(=)
(a+b)*a
(a*b*)*
=(a+b)*
(a+b)*a(a+b)*a(a+b)*
(b*ab*ab*ab*)*

PASCAL unsigned numbers. d={0,1,…,9}

dd*(ε +.dd*)(ε+E(+ + – + ε)dd*)
aΣ*a+b Σ*b+a+b

Defn: E=F ⇔L(E)= L(F)

 ∅*= ε (E*F*)*=(E+F)* E∅=∅E=∅  E(FG)=(EF)G E(F+G)=EF+EG Eε=εE=E

L

{ , }{ } a b a

({ }{ })

a b

{ , }

a b =

{ : has 2 a's} w w

{

w : # a's divisable by 3} { : ?? } w { : begins & ends same} w w

C SC 473 Automata, Grammars & Languages 26

Nondeterminism

Real computing devices are deterministic: the current

configuration and instruction determines the next

configuration. The relation is a function.
Why the concept of nondeterminism?

 Provides powerful, economical descriptive ability  Provides a way to specify languages without over-specifying and

complex handling of cases

 Can be algorithmically converted to a deterministic description (at

the sacrifice of some economy and with added complexity)

 Generalization of determinism

Ex: abab occurs somewhere in w: …abab…
a

b a b a,b a,b a,b a b a

C SC 473 Automata, Grammars & Languages 27

Nondeterminism (contʼd)

Ex: w∈Σ* has penultimate symbol b: w = …b?
Ex: w∈Σ* has ≥ 2 a’s: w = … a … a …

b b a,b a b a,b a a a a,b a,b a,b

SLIDE 10

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 10

C SC 473 Automata, Grammars & Languages 28

ε-Moves Can Be Useful

SNOBOL arithmetic constants (no floating E)

 Use to specify “optional characters” like Unix command line [opt]

ε

−

d d=digit d d ε

d

C SC 473 Automata, Grammars & Languages 29

Nondeterministic Finite Automaton

Defn 1.5: A nondeterministic finite automaton is a 5-tuple



is a finite set, the states



is a finite set, the alphabet



transition function



is the start state



is the set of accepting (final) states

Ex:

( , , , )

,

M Q q

F

=
F

Q

Q

: ( { }) ( ) Q Q

P
q

Q

1

( , , , )

,

M Q q

F

=
1

{ , } Q q q =

1

{ , } { } a b F q = =

1 1 1 1 1

( , ) { } ( , ) { , } ( , ) { } ( , ) { } q a q q b q q q a q q b q

=

= = =

1

M

b a,b a,b

q

1

q

C SC 473 Automata, Grammars & Languages 30

DFA vs NFA

DFA δ
For each state q and input symbol

a, there is exactly one choice of new state (or no transition is defined at all). Each transition “consumes” an input symbol

Special case of NFA!
NFA δ
There may be multiple choices for

the same input symbol

There may be ε-moves that do not

“consume” an input character

There can be “chains” of ε-moves
ε-moves can create even more

choice for the next input character

q a b q a a b b ε ε

SLIDE 11

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 11

C SC 473 Automata, Grammars & Languages 31

How NFA Compute

Given a NFA
Defn: configuration –
Defn: yields in one step (or moves) relation

between configurations

Defn: yields =

 Means “COULD move in zero or more steps to”

Defn: w is recognized (accepted) by M ⇔

 Same as before, but has the meaning “if there exists some

sequence of moves from the start config to some accepting config”

( , , , )

,

M Q q

F

=
(

q,aw) ( q ,w)

q

( q,a)

( , ) q w Q

!
!

( f F)( q0,w) ( f,) ( q,w) ( q ,w)

q

( q,)

(ε-move)

C SC 473 Automata, Grammars & Languages 32

How NFA Compute (contʼd)

Defn: The language recognized (accepted) by M is
Ex: In NFA

 This provides no “evidence” that aabbba is accepted (or not)  However, also via a separate computation sequence:  And so aabbba is recognized!

L (M) = { w : M accepts w} = { w :( f F)( q0,w)

(

f,)

}

( q0,aabbba) ( q0,)

1

M

( q0,aabbba) ( q1,)

C SC 473 Automata, Grammars & Languages 33

“Tree” of Computations

Ex: NFA M1

( , ) q aabbba ( , ) q abbba ( , ) q bbba

1

( , ) q bba ( , ) q bba ( , ) q ba

1

( , ) q ba ( , ) q a

1

( , ) q a ( , ) q

1

( , ) q

X

null “evidence” ∃ accepting Computation ⇒ w∈L(M1)

1

q F

=

SLIDE 12

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 12

C SC 473 Automata, Grammars & Languages 34

Computation Tree: Example

Ex: L = {w: w begins & ends same }

2 a

3 4

a,b

1 a,b b b a ( , )

2

( , )

3

( , )

1 ababa

( , )

2 baba

( , )

2 aba

( , )

2 ba

( , )

3 ba

( , )

2 a

X

3∈F

( , )

1 abab

( , )

2 bab

( , )

2 ab

( , )

2 b

( , )

3 b

( , )

2

X X

reject abab ¬∃ path to F accept ababa ∃ some path to F

C SC 473 Automata, Grammars & Languages 35

Example with ε-Moves

String length a multiple of 2 or 3

2 4 3 ε ε a 5 6 a a a a ( , )

0 aaa

( , )

2 aaa

( , )

4 aaa

( , )

3 aa

( , )

2 a

( , )

5 aa

( , )

6 a

( , )

3

( , )

4 !

X

4∈F ⇒ Accept aaa ← ε-moves

C SC 473 Automata, Grammars & Languages 36

Example with ε-Moves

a*b*

( , )

0 aab

( , )

1 aab

( , )

0 ab

( , )

1 ab

( , )

0 b !

X

1∈F ⇒ Accept aab

X

( , )

1 b

( , )

1

ε a 1 b ε-moves “consume” no input symbols

SLIDE 13

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 13

C SC 473 Automata, Grammars & Languages 37

Equivalence of NFA to DFA

There is an algorithm to convert any NFA into a DFA

 We show basic idea assuming NFA has no ε-moves  Then (later) modify the construction for NFAs with ε-moves

Ex: L = {x : last symbol of x appeared previously } Σ={a,b}
Idea: given input string, keep track of all possible reached

states after reading each letter. At end of input, see if a final state is among those reached p q r s a b a,b a,b a,b a b N0:

C SC 473 Automata, Grammars & Languages 38

Equivalence of NFA and DFA (contʼd)

Computation paths through NFA N0 on w = abba

p p p p p r r r q q q q r r s s q a a b b a a b a b b a b b b a a

s F

C SC 473 Automata, Grammars & Languages

39

Equivalence of NFA and DFA (contʼd)

Idea: keep a list of all possible states reachable by each

prefix of w (“parallel worlds”). For NFA N0:

{ } { , { , { , { , } , , , } , , } } { } { , , , } ( ) since { , , , }

a b b a abba

p p p p p q q q q r r r s s p p q r s abba L N p q r s F

SLIDE 14

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 14

C SC 473 Automata, Grammars & Languages 40

Equivalence of NFA and DFA (contʼd)

Equivalent DFA M will have:

 State set P (Q)  Alphabet Σ  Start state “set” {q0}  Accepting states {X ⊆ Q : X ∩ F ≠ ∅}  Deterministic transition function δ′ : P (Q) × Σ → P (Q)

Ex: For NFA N0:

({ , }, ) { , , } ({ , }, ) { , , } ({ }, ) { , } ({ }, ) { , } p q a p q s p q b p q r p a p q p b p r

=
=
=
=

…

C SC 473 Automata, Grammars & Languages 41

Equivalence of NFA and DFA (contʼd)

Thm: [Rabin-Scott Construction]. Let L = L(N) for some NFA

N with no ε-moves. There is an algorithm to constuct a DFA M equivalent to N, i.e. with L(M) = L(N). Pf: Given N we construct a DFA M and then verify that it recognizes the same set as N. Construction: Given NFA construct where

( , , , , ) N Q s F

=
( , ,

, , ) M Q s F

=
Q

= P( Q),

s

= { s},

F

= { X

Q : X F }

and :

Q
Q

is defined as: (S Q,a )

(

S,a) = { q Q :(p S)q (p,a)}

C SC 473 Automata, Grammars & Languages 42

Equivalence of NFA and DFA (contʼd)

Picture of δ′
(

S,a) =

S

a a a b a b

S S

S Q S Q

SLIDE 15

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 15

C SC 473 Automata, Grammars & Languages 43

Equivalence of NFA and DFA (contʼd)

Verification: Show (1) M is a DFA and (2) L(M)=L(N)

(1) δ′ is a function by the construction, and Q′ is finite: |Q′ | = 2|Q| . So M is a DFA. (2) To show equivalence we prove the

Lemma:

Pf: By induction on the length of the input string w. Base |w|=0. Step Suppose (IH) the lemma is true Let To show:

(p,w) N

(

q,) Q. q Q ({p},w)

M

(

Q,)

(p,) N

(

q,) p = q ({p},)

M

({

q},) . | | . w w k

| w |

= k + 1 ,w = ua,| u |= k. (p,ua) N

(

q,) Q. q Q ({p},ua)

M

(

Q,)

C SC 473 Automata, Grammars & Languages 44

Equivalence of NFA and DFA (contʼd)

⇒. Assume Then ∃ state r with

and Then By (IH) () By construction of M Let Then Using this with () results in: So

(p,ua) N

(

q,).

(p,ua) N

(

r,a) N ( q,) (p,u) N

(

r,).

R. r R ({p},u)

M

(R,)

( , ). q r a

( , ).

q R a

( , ).

Q R a

=

Q. q Q (R,a)

M

(

Q,). Q. q Q ({p},ua)

M

(R,a)

M

(

Q,). Q. q Q ({p},ua)M

(

Q,).

C SC 473 Automata, Grammars & Languages 45

Equivalence of NFA and DFA (contʼd)

⇐. Assume

Then ∃ state R with So (1) By construction and (2) Since we have from (IH) (3) Combining (2) & (3): So (p,ua) N

(

q,).

(p,ua) N

(

r,a) N ( q,)

({p},u)

M

(R,)

. ( , ) r R q r a

( , )

. R a Q

=

({p},ua)

M

(R,a)

M

(

Q,). Q. q Q ({p},ua)M

(

Q,).

( r,a) N

(

q,). (p,u) N

(

r,)

SLIDE 16

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 16

C SC 473 Automata, Grammars & Languages 46

Equivalence of NFA and DFA (contʼd)

We now finish the verification proof. Let

From the Lemma That is, for some for some ( s ,w)M

(

Q,) Q. f Q ({ s},w)M

(

Q,)

. f F

(

s,w) N

(

f,) ( s ,w)M

(

Q,) Q F . ( s,w) N

(

f,) f F

.

Q F

( )

( ). L M L N

=
C SC 473 Automata, Grammars & Languages

47

Example: ε-Free NFA → DFA

Consider the previous NFA

( , , , )

,{ }

N Q

p s

=
( ( ), ,

,{ )

},

M Q

p F

=

P

{p} {p,q} {p,r} {p,q,s} {p,q,r} {p,r,s} {p,q,r,s} a a b b a b a b a b a,b b a

C SC 473 Automata, Grammars & Languages 48

ε-closure(R) = E(R) for a set of

states R

a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ε ε a ε ε ε a ε b ε ε ε c b d

closure({4, 5})

=E({4, 5}) ={1, 2, 3, 4, 5, 6,7, 8}

closure (R) = { q Q :(r R) ( r,) ( q,)}

For R ⊆ Q the ε-closure, E(R) of R is:

R

NFA with ε-Moves

SLIDE 17

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 17

C SC 473 Automata, Grammars & Languages 49

ε-closure of a set of states

Coalesce all nodes reachable from {4,5} by ε-moves:

a 9 10 11 12 13 14 15 a a b c b d {1,2,3,4,5,6,7,8} Note: still an NFA

E({4,5}) E({9}) E({13})

a a Etc. a

closure({4, 5})

=E({4, 5}) ={1, 2, 3, 4, 5, 6,7, 8}

C SC 473 Automata, Grammars & Languages 50

Conversion: NFA → DFA

Thm: There is an algorithm to convert any NFA to an

equivalent DFA .

Pf: Construction: Given NFA

construct new NFA where Verification. Pf: By induction on |w|

M = ( 2

Q,,

, s , F ) ( , , , , ) N Q s F

=
F

= { S Q | S F }

: 2

Q 2 Q

s

= E ( s)

(

S,a) = { E (p) | p ( q,a) for some q S} ( q,w)

N (p,) (E

( q),w)

M (

P,) for some set P containing p

C SC 473 Automata, Grammars & Languages 51

Conversion: NFA → DFA

Thm 1.39: [Rabin-Scott Theorem]: There is an algorithm

to convert any NFA into an equivalent DFA.

Corollary 1.40: A language is regular ⇔ some NFA

recognizes it.

Ex: Start with an NFA N1 as follows:

ε

−

1 2 3 4 b b b d ε

1

N = {b,,d}

SLIDE 18

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 18

C SC 473 Automata, Grammars & Languages 52

Conversion: NFA → DFA

Ex:

ε

−

1 2 3 4 b b b d ε

1

N

E ( 1 ) = { 1 } E ( 2 ) = { 1 ,2 ,3 } E ( 3 ) = { 1 ,3 } E ( 4 ) = { 4 } 1

b

2 2

b

2 3

b

4 4

b

1
2
3

3

4
1

d

2

d

3

d

4

d

3 Useful summary

C SC 473 Automata, Grammars & Languages 53

Conversion: NFA → DFA

Ex:

ε

−

1 2 3 4 b b b d ε

1

N

s

= E ( 1 ) = { 1 }

(

s ,b) = E ( 2 ) = { 1 ,2 ,3 }

(

s ,) =

(

s ,d) =

({

1 ,2 ,3 },b) = E ( 1 ) E ( 2 ) E ( 3 ) E ( 4 ) = { 1 ,2 ,3 ,4 }

({

1 ,2 ,3 },) = E ( 3 ) = { 1 ,3 }

({

1 ,2 ,3 },d) =

C SC 473 Automata, Grammars & Languages 54

Ex: NFA → DFA (contʼd)

Ex:

ε

−

1 2 3 4 b b b d ε

1

N

({

1 ,3 },b) = E ( 2 ) E ( 4 ) = { 1 ,2 ,3 ,4 }

({

1 ,3 },) =

({

1 ,3 },d) =

({

1 ,2 ,3 ,4 },b) = E ( 2 ) E ( 4 ) = { 1 ,2 ,3 ,4 }

({

1 ,2 ,3 ,4 },) = E ( 3 ) = { 1 ,3 }

({

1 ,2 ,3 ,4 },d) = E ( 3 ) = { 1 ,3 }

SLIDE 19

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 19

C SC 473 Automata, Grammars & Languages 55

Conversion: NFA → DFA (contʼd)

1

M

−

1 123 13 1234

b b b d

−

b

C SC 473 Automata, Grammars & Languages 56

Regular Expression → NFA

Thm 1.55: There is an algorithm that, given a regular

expression E, constructs a NFA N such that L(E) = L(M). Pf: Induction on the # of operator symbols in E. Base: E = ε ∅ a∈Σ Step: Assume (IH) the result is true of all expressions with ≤ operator symbols (+,⋅,*). Let E have k+1 ops. Three cases: Case E = (E1+E2). By IH, ∃ FA M1 , M2 with L(E1) = L(M1) and L(E2) = L(M2). Construct the following NFA M. a

C SC 473 Automata, Grammars & Languages 57

1

M

2

M

ε ε

1

F

2

F M

1 2

F F F =

1

2

( ) ( ) ( ) L M L M L M =

Case +

SLIDE 20

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 20

C SC 473 Automata, Grammars & Languages 58

Regular Expression → NFA (cont”d)

Case E = (E1⋅E2). By IH, ∃ FA M1 , M2 with L(E1) = L(M1)

and L(E2) = L(M2). Construct the following NFA M.

C SC 473 Automata, Grammars & Languages 59

2

M

ε

M

2

F F =

1 2

( ) ( ) ( ) L M L M L M =

i

Case •

2

F

1

F

1

M

ε

Unmark final states in M1

C SC 473 Automata, Grammars & Languages 60

Regular Expression → NFA (cont”d)

Case E = (E1)*. By IH, ∃ FA M1 with L(E1) = L(M1).

Construct the following NFA M.

SLIDE 21

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 21

C SC 473 Automata, Grammars & Languages 61

s

M

1

{ } F F s =

1

( ) ( ) L M L M

=

Case *

1

F

1

M

ε ε ε

QED

C SC 473 Automata, Grammars & Languages 62

ε b ε ε ε a ε a ε b a a

Example: Reg. Exp.→NFA

(b+aa)*

Not very economical

C SC 473 Automata, Grammars & Languages 63

Regular Expressions—Applications

Regexp used in various development tools

 qed – interactive text editor. 1st version Lampson & Deutsch 1967 ◆ Regexp added by Ken Thompson, Bell Labs, ca. 1968  Regexp compiled into NFA in machine code  Rabin-Scott idea used to scan “on the fly”  One of the first software patents ◆ Offspring ed by Ken for Unix ◆ Many others followed: em, vi / ex, sam, qedx, …  grep, egrep - pattern search in a file  shell – command line interpreter  lex – lexical analyzer generator  sed – non-interactive stream editor  awk – pattern scanning and processing language  perl – pattern-driven programming language

SLIDE 22

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 22

C SC 473 Automata, Grammars & Languages 64

Applications (contʼd)

Regular expressions = “patterns”

meaning awk regexp matches >=1 r r+ matches >=0 r r* matches 0 or 1 r r? matches r then s rs matches r or s r|s match literal “c” \c match begin/end line ^ $ match any char

.

group exprs (s) character list [abc…] negated char list [^abc…]

C SC 473 Automata, Grammars & Languages 65

Applications--Examples

?[0-9]+

nonempty digit strings, optional sign [^0-9] any char except digit \[.\] reference citations in a paper g/^[ ]$/d delete blank lines g/[ ]+/d delete lines with a blank Ex: match is always (1) leftmost and (2) longest file: abcddddef vi: s/d*/x/ → xabcddddef s/d+/x/ → abcxef Ex: csh: sort roll[1-5] | egrep “C SC|MATH” | pr

C SC 473 Automata, Grammars & Languages 66

Applications--Examples

Ex: traditional spelling mnemonic

“i before e, except after c,
r when pronounced ʻaʼ,

as in ʻneighborʼ and ʻweighʼ

-except for ʻweirdʼ examples.”
grep “[^c]ei” /usr/share/dict/words > foo

cat foo abseil Aeneid ageing Alamein albeit atheist Boeing Budweiser caffein canoeist deice deictic dilettanteism dreidl ...

if you think this spelling rule is sufficient, you will be deficient,

inefficient, unscientific and far from omniscient

SLIDE 23

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 23

C SC 473 Automata, Grammars & Languages 67

Applications--Examples

Ex: lex – generates a lexical analyzer yylex(). Example: wordcount (wc) %{ int nchar, nword, nline; %} %% \n { nline++; nchar++; } [^ \t\n]+ { nword++, nchar += yyleng;} // yyleng = length of matched string . { nchar++;} %% int main(void) { yylex(); // invoke generated lexer printf("%d\t%d\t%d\n", nchar, nword, nline); return 0; }

C SC 473 Automata, Grammars & Languages 68

L Regular ⇔ L Denoted by a Reg. Expr.

Weʼve defined “regular” as meaning: recognized by a DFA

(equiv. to rec. by an NFA)

This equivalence result is known as “Kleeneʼs Theorem”
Weʼve already shown the ⇐ direction—we constructed an

NFA from a regular expression (Using Rabin-Scott we could convert this NFA to a DFA.)

Now we show the ⇒ direction: given a DFA M construct

a regular expr. E with L(M) = L(E).

Thm (Kleene): There is an algorithm that, given a DFA M ,

computes a regular expression E such that L(M) = L(E).

Pf: Given the graph of the DFA, use the “node elimination

algorithm” to gradually eliminate all nodes in favor of expressions on the edges of the graph.

C SC 473 Automata, Grammars & Languages 69

B C 1 1 A 1 S a ε ε B C 1 1 A 1 B 1 00 1 ε ε

Kleeneʼs Thm: use “Generalized NFA”

A S a 01 A S a 1(1+01)00+0 ε ε S a ε(1(1+01)00+0)*ε

E =(1(1+01)00+0)

1. Add init. S, & final a

2. Elim. C
3. Elim. B
4. Elim. A

Order: CBA

SLIDE 24

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 24

C SC 473 Automata, Grammars & Languages 70

Ex: Node Elimination Algorithm

B A

b b a b a

C B

C

A

b b a b a

Add ε-moves:

S

ε a ε

C SC 473 Automata, Grammars & Languages 71 B

C b b

ba*a

S

ε a ε

B

bb bba*a

S

ε a b (bb+bba*a)*b

S

a

Elim. A:
Elim. C:
Elim. B:

ACB

Ex: Node Elimination Algorithm (ACB)

A AC

C SC 473 Automata, Grammars & Languages 72 B

C

A

b b a b a

S

ε a ε

Ex: Node Elimination: other orders

SLIDE 25

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 25

C SC 473 Automata, Grammars & Languages 73 B

b

S

ε a

B

bb bba*a

S

ε a b (bb+bba*a)*b

S

a

Elim. C:
Elim. A:
Elim. B:

CAB=ACB

Ex: Other elimination orders (CAB)

bb

A

bb a a

CA= AC (above) C

C SC 473 Automata, Grammars & Languages 74 B

b

S

ε a

A S

a (bb)*b (bb)*b+(bb)*bba*a(bb)*b

S

a

Elim. C:
Elim. B:
Elim. A:

CBA

Ex: Other elimination orders (CBA)

bb

A

bb a a

CB C

(bb)*bb a a(bb)*b

C SC 473 Automata, Grammars & Languages 75 A S

a b(bb+ba*ab)*

S

a

Elim. B:
Elim. A:

BAC=ABC

Ex: Other elimination orders (BAC)

BA =AB B

a

C

b bb ε b ab

S

a

C

b bb ba*ab ε

Elim. C:

SLIDE 26

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 26

C SC 473 Automata, Grammars & Languages 76

Ex: All elimination orders equiv (BAC = ACB)

BAC

b(bb+ba*ab)*

ACB

(bb+bba*a)*b Easy to prove by induction: for any expression E, b(Eb)*=(bE)*b. Using this identity: b(bb+ba*ab)* = b[ (b+ba*a) b]* = [b (b+ba*a) ]*b = (bb+bba*a)*b Further regular expression simplication is possible:

b(bb+baab)=b[b(b+aab)]=b[b(ε+aa)b] =b[bab]

Good exercise: show results of all other elimination orders are equivalent to these, using regular expression algebra

C SC 473 Automata, Grammars & Languages 77

Algebra of Regular Expressions

∃ an algebra for symplifying regular expressions
Can use this algebra to construct RegExps from FSA

r+s = s+r (r+s)+t=r+(s+t) r+r=r r+∅=r (rs)t = r(st) rε = εr = r r∅ = ∅r = ∅ r(s+t) = rs + rt (r+s)t = rt + st ∅* = ε (r) = r* r* = r + r* r* = ε + rr* (rs)* = (r+s)*

C SC 473 Automata, Grammars & Languages 78

Solving Regular Expr Equations

Can solve “linear equations” with regexp variables

X = aX + b =a(aX + b) + b = a2X + ab + b = a2 (aX + b) + ab + b = a3X + a2b + ab + b … X = ab Check: a[ab] + b = aab + b = (aa+ε)b = a*b Ex: X

b a

Y X = aX + bY Y = ε  X = a*b

SLIDE 27

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 27

C SC 473 Automata, Grammars & Languages 79

Solving RegExp Equations (contʼd)

Ex: NFA  RegExp

B C 1 1 A 1

1 1 1 _ elim. 1 0 11 0 11 0 _ elim. ____ ( 11 0 )0 11 0 ( 11 0 )0 [ 11 0 ( 11 0 )0 ] [ 11 0 ( 11 0 )0 ] Simplify using reg. algebra: 11 0 ( 11 0 )0 [ 11 0 ( 11 A A B B B C C A B B B C A A C C A C C C A A A A A

=

+ + = + = + = = + + = + = = + + = + + = + + = + )] ( 11 0 )0 (( 11 0 ) 0 ) A

=
=

Gauss-Jordan elimination & back-substitution

C SC 473 Automata, Grammars & Languages 80

Ex: Node Elimination Example via Algebra

Want B. C is accept state.
Elim. A:
Elim B:
Elim C:
∴

B A

b b a b a

C

A aA aB B bC C bA bB

=

+ = = + +

* A a aB =

* B bC C ba aB bB

=

= + +

B bC = * ( * ) C ba abC bbC ba ab bb C

=

+ + = + + ( * ) * C ba ab bb = + ( * ) * B bC b ba ab bb = = + Simplifies to = ( * ) * B b ba b

C SC 473 Automata, Grammars & Languages 81

Closure Properties

A class of languages is said to be closed under an
peration if applying that operation to members of the

class results in a language that is again a member of the

class. Example: the regular languages are closed under

the operations of union, concatenation and Kleene star.

Thm: The regular languages are closed under intersection

and complementation. Pf: Complementation. Let L = L(M) where is a DFA. Then the FA is also deterministic, and So w leads to a non- accepting state in ⇔ w leads to an accepting state in So

( , , , , ) M Q s F

=
( , , , , )

M Q s F

=
(

s,w) M

(

q,) ( s,w) M

(

q,). . M M ( ) ( ). L M L M

=

SLIDE 28

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 28

C SC 473 Automata, Grammars & Languages 82

Closure Properties (contʼd)

Intersection. Let be regular. By DeMorganʼs law

Since the regular languages are closed under complementation and union, the result follows.

1 2

, L L

1 2 1 2

( ). L L L L

=
C SC 473 Automata, Grammars & Languages

83

Closure Properties (contʼd)

Another proof of closure under ∩ illustrates the technique

called “cross-product construction”. See Sipser text, Theorem 1.25.

Thm: The class of regular languages is closed under the

intersection operation. Pf: Assume and , where the automata are deterministic.

Construction. Construct a “cross-product machine” M as

follows: where the transition function is defined by: Machine M simulates the two given machines “in parallel”, keeping each machine state in one component of

1 1 1 1 1 1 1 1

( ), ( , , , , ) L L M M Q s F

=

=

2

2 2 2 2 2 2 2

( ), ( , , , , ) L L M M Q s F

=

=

1

2 1 2 1 2 1, 2 1 2

( , , ,( ), ) M Q Q s s F F

=
1

2 1 2 1 1 2 2

(( , ), ) ( ( , ), ( , )). q q a q a q a

=

1 2

( , ). q q

C SC 473 Automata, Grammars & Languages 84

Closure Properties (contʼd)

Verification By an easy induction on |x|, can show that Therefore, for a pair of final states This says that i.e., that

(( q1,q2),x) M

((p1, p2),)

( q1,x)M1

(p1,) (

q2,x)M2

(p2,)

(( s1,s2),x)M

((

f

1,f 2),)

( s1,x)M1

(

f

1,) (

s2,x)M2

(

f

2,) 1 1 2 2

, f F f F

1

2

( ) ( ) ( ) x L M x L M x L M

1

2

( ) ( ) ( ). L M L M L M =

SLIDE 29

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 29

C SC 473 Automata, Grammars & Languages 85

Closure Properties (contʼd)

Defn: A homomorphism h is a function that maps each

symbol of to a string over some alphabet Δ, i.e.,

The homomorphism is extended to operate on strings

character-by-character, i.e.,

It is further extended to languages element-wise, i.e.,
Thm: If L is regular and h is a homomorphism, then h(L) is

regular. Pf: Assume L is recognized by a DFA

1 2

{ , , , }

n

a a a = …

1 1 2 2

( ) , ( ) , , ( )

n n

h a w h a w h a w = = = …

1 2 1 2

( ) ( )( ) ( ).

n n

hc c c hc hc hc = … … ( ) {( ): }. h L hw w L =

( , , , , ).

M Q s F

=
C SC 473 Automata, Grammars & Languages

86

Closure Properties (contʼd)

Construction: Construct the machine

where for each transition in M: put into Mh the transition An easy induction establishes that from which it follows that

( , , , , )

h h

M Q s F

=
a

p q

h(a)

p q ( s,w) M

(

q,) ( s,h ( w)) M h

(

q,),

( ) ( ( )).

h

L M h L M =

Note: GNFA

C SC 473 Automata, Grammars & Languages 87

What is Not Regular?

FA have a very limited computing ability. They cannot, for

example, recognized strings of well-nested parentheses,

r well-formed arithmetic expressions, or even the

language of strings of the form w#w, having two copies of the same substring.

How can we show some languages are not regular? We

will give a property that all regular languages must have (called the pumping property). Then, to show that a language L is not regular, we argue that it lacks this pumping property.

SLIDE 30

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 30

C SC 473 Automata, Grammars & Languages 88

Pumping Lemma

Thm [Pumping Lemma for Regular Languages]. Suppose

that L is an infinite¹ regular language. Then

¹All finite languages are regular, so only infinite languages are

f interest.

( )( )[ ( , , )( ( ) )]

i

p w w L w p x y z w xyz y xy p i xy z L

=
C SC 473 Automata, Grammars & Languages

89

Pumping Lemma (English)

Thm [Pumping Lemma for Regular Languages]. Suppose

that L is an infinite¹ regular language. Then there is some number p (called the pumping length) such that: if w is any string in L with |w| ≥ p, then w can be factored into 3 substrings, w = xyz, that satisfy the following 3 conditions: (i) y ≠ ε [y is not empty] (ii) |xy| ≤ p [the prefix and pumped part are short] (iii) for every i ≥ 0, xyiz ∈ L [“pumped up” and “pumped down” (i = 0) versions of the string must also be in L]

¹All finite languages are regular, so only infinite languages are

f interest.

C SC 473 Automata, Grammars & Languages 90

Pumping Lemma (contʼd)

Pf: Let be a DFA recognizing L and

let p be the number of its states. Let be an input string of length n where n ≥ p. Let be the sequence of states that M enters while processing w so that for This state sequence has length n+1 ≥ p+1. Among the first p+1 states of this sequence, at least 2 must be the same state [“pigeonhole principle”]. Call the first of these 2 and the second . Because

ccurs among the first p+1 places in the

sequence we have that k ≤ p+1. Define the following substrings of w: ( , , , , ) M Q s F

=
1 2

n

w a a a =

1

2 1

, , ,

n n

r r r r +

1

( , )

i i i

r r a

+

= 1 . i n

j

r

k j

r r =

1 2 1

, , ,

n n

r r r r +

1

1 1

, ,

j j k k n

x a a y a a z a a

=

= =

k

j

r r =

SLIDE 31

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 31

C SC 473 Automata, Grammars & Languages 91

a rk rn+1 r1

1 1 j

x a a =

1

j k

y a a =

=rj

k n

z a a =

Pumping Lemma (contʼd)
Picture:
From the picture, we see that there is an accepting path

from to a final state for all the strings of the form . Also, since it must be that Furthermore, so .

1

s r =

1 n

r + ,

i

xy z i j k

.

y

1

k p

+

. xy p

C SC 473 Automata, Grammars & Languages

92

Non-regular Examples

Ex: is not regular.

Pf: By contradiction. Suppose L is regular. Then by the Pumping Lemma, Then it follows that is a perfect square. This is impossible. For suppose for a so large that Then Hence falls “in between” perfect squares—a contradiction.

2

{ : }

k

L a k =

(

, , ) , ( ), [( ) ( ) .

p q r p q n r

x y z x a y a q z a n a a a L

=

= > =

(

) n p q n r

+
+

2

p r q n k + +

=

k 2 1 . k q + > ( 1 ) p r q n + +

+

2 2 2

2 1 ( 1 ). k q k k k = + < + + = +

( 1 ) p r q n + +

+
C SC 473 Automata, Grammars & Languages

93

Non-regular Examples (contʼd)

Ex: is not regular.

Pf: By contradiction. Suppose L is regular. Then by closure properties of the regular languages is regular. Now We show cannot be regular, which provides a contradiction. If is regular, then there are substrings with such that Case 1. y is entirely in the aʼs. Assume it is in the aʼs before the 2 bʼs (The other subcase is symmetric). Then and But then where This is a contradiction.

{ : { , }}

R

L w w w a b =

*

* 1

L L a bba =

1

{ : }.

n n

L

a bba n =

1

L

1

L

y

, ,

x y z

1

.

n

n xy z L

,

, ( )

p q r s

x a y a z a bba q = = = > . p q r s + + =

p r s

a a bba L

.

p r s + <

SLIDE 32

CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 32

C SC 473 Automata, Grammars & Languages 94

Non-regular Examples (contʼd)

Case 2. y contains a b. Then has more than 2

bʼs, and so This is a contradiction. Contradictions in all cases ⇒ contradiction to the assumption that is regular. So is not regular.

Ex: is not regular. See Text,

Example 1.73.

Ex:

is not regular. Pf: Suppose B is regular. Then so is as is its homomorphic image Contradiction.

1

L

3

xy z

3 1.

xy z L

1

L {

: }

n n

L a b

n

=

{

: is a well-nested string of parentheses }

B

w w

=

{( ) : }

n n

n =

( )

B B =

{

: }.

n n

a b n

C SC 473 Automata, Grammars & Languages

95

Decision Problems

For a property/predicate P the decision problem for P is:



Given: x



Question: Is P(x) true?

Ex: Given DFA M, is it true that L(M) = ∅?
Thm: For DFA M all the following decision problems are

solvable, i.e. there exists an algorithm to decide the question for any input: Given Question

1. M,w

w ∈ L(M) ?

2. M

L(M) = ∅ ?

3. M L(M) = Σ* ?
4. M, M′

L(M) ⊆ L(M′ ) ?

5. M, M′

L(M) = L(M′ ) ?

C SC 473 Automata, Grammars & Languages 96

Decision Problems (contʼd)

Pf: Assume given DFAs for inputs. (1) Trace w through M. “Yes” if leads from s to some q ∈ F (2) “Yes” if there is some q ∈ F reachable from s (3) Convert (4) (5) Use (4) twice . ( ) ( ) M M L M L M

=
=

1 2 2 1

( ) ( ) ( ) ( ) L M L M L M L M

=

2

( ) L M

1